14.3 Phase change and latent heat  (Page 3/9)

Calculate final temperature from phase change: cooling soda with ice cubes

Three ice cubes are used to chill a soda at $\text{20º}\text{C}$ with mass $m_{\mathrm{soda}}=\text{0.25 }\text{kg}$ . The ice is at $\mathrm{0º}\text{C}$ and each ice cube has a mass of 6.0 g. Assume that the soda is kept in a foam container so that heat loss can be ignored. Assume the soda has the same heat capacity as water. Find the final temperature when all ice has melted.

Strategy

The ice cubes are at the melting temperature of $\mathrm{0º}\text{C}$ . Heat is transferred from the soda to the ice for melting. Melting of ice occurs in two steps: first the phase change occurs and solid (ice) transforms into liquid water at the melting temperature, then the temperature of this water rises. Melting yields water at $\mathrm{0º}\text{C}$ , so more heat is transferred from the soda to this water until the water plus soda system reaches thermal equilibrium,

The heat transferred to the ice is $Q_{\text{ice}}=m_{\text{ice}}{L}_{\text{f}}+m_{\text{ice}}{c}_{\text{W}}(T_{\text{f}}-\mathrm{0º}\text{C})$ . The heat given off by the soda is $Q_{\text{soda}}=m_{\text{soda}}{c}_{\text{W}}(T_{\text{f}}-\text{20º}\text{C})$ . Since no heat is lost, $Q_{\text{ice}}=-Q_{\text{soda}}$ , so that

Bring all terms involving $T_{\text{f}}$ on the left-hand-side and all other terms on the right-hand-side. Solve for the unknown quantity $T_{\text{f}}$ :

Solution

1. Identify the known quantities. The mass of ice is $m_{\text{ice}}=3\text{×}\mathrm{6.0 }\text{g}=0\text{.}\text{018}\text{ kg}$ and the mass of soda is $m_{\text{soda}}=0\text{.}\text{25 }\text{kg}$ .
2. Calculate the terms in the numerator:
   $m_{\text{soda}}{c}_{\mathrm{W}}\left(\text{20º}\text{C}\right)=\left(0\text{.25 kg}\right)\left(\text{4186 J/kg}\cdot º\text{C}\right)\left(\text{20º}\text{C}\right)=\text{20,930 J}$

   and

   $m_{\text{ice}}{L}_{\text{f}}=\left(0\text{.}\text{018 }\text{kg}\right)\left(\text{334,000}\text{ J/kg}\right)\text{=6012 }\text{J.}$
3. Calculate the denominator:
   $\left(m_{\text{soda}}+m_{\text{ice}}\right)c_{\text{W}}=\left(0\text{.}\text{25 }\text{kg + 0}\text{.}\text{018 kg}\right)\left(\text{4186 K/}(\text{kg}\cdot º\text{C}\right)\text{=1122 J/º}\text{C}\text{.}$
4. Calculate the final temperature:
   $T_{\text{f}}=\frac{\text{20},\text{930 J}-\text{6012 J}}{\text{1122 J/º}\text{C}}=\text{13º}\text{C.}$

Discussion

This example illustrates the enormous energies involved during a phase change. The mass of ice is about 7 percent the mass of water but leads to a noticeable change in the temperature of soda. Although we assumed that the ice was at the freezing temperature, this is incorrect: the typical temperature is $-\mathrm{6º}\text{C}$ . However, this correction gives a final temperature that is essentially identical to the result we found. Can you explain why?

We have seen that vaporization requires heat transfer to a liquid from the surroundings, so that energy is released by the surroundings. Condensation is the reverse process, increasing the temperature of the surroundings. This increase may seem surprising, since we associate condensation with cold objects—the glass in the figure, for example. However, energy must be removed from the condensing molecules to make a vapor condense. The energy is exactly the same as that required to make the phase change in the other direction, from liquid to vapor, and so it can be calculated from $Q={\text{mL}}_{\text{v}}$ .

Real-world application

Energy is also released when a liquid freezes. This phenomenon is used by fruit growers in Florida to protect oranges when the temperature is close to the freezing point $\left(\mathrm{0º}\text{C}\right)$ . Growers spray water on the plants in orchards so that the water freezes and heat is released to the growing oranges on the trees. This prevents the temperature inside the orange from dropping below freezing, which would damage the fruit.